Timeline for $p_n(x,y)=\sum_{i=0}^{n-1}x^{n-1-i}y^{i}$ is always an integer
Current License: CC BY-SA 4.0
7 events
when toggle format | what | by | license | comment | |
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Jan 19, 2019 at 23:09 | vote | accept | jack | ||
Nov 19, 2018 at 20:08 | comment | added | Vlad Matei | If you vary $m$ also you end up with an intersection of two surfaces if $l=2$, and at least four if $l\geq 3$. Bombieri Lang conjecture gives finitness of rational points so this why I believe it should be rare, or even impossible if you have more than four surfaces. | |
Nov 19, 2018 at 20:05 | comment | added | Vlad Matei | I would bet that yes. Here is a way to go about it, though its more heuristic than a proof. From above you get that $ab=\sqrt[l]{m}$, $m\in\mathbb{Q}$ is not an $l$ power where $l|k$ and $a+b=p+q\sqrt[l]{m}$. Suppose $l\geq2$. Then expressing $\cfrac{a^k-b^k}{a-b}$ and $\cfrac{a^{k+1}-b^{k+1}}{a-b}$ in terms of $a+b$ and $ab$ and using the fact that $\sqrt[l]{m},\ldots, \sqrt[l]{m^{l-1}}$ are algebraically independent over $\mathbb{Q}$ you get two curves on which $(p,q)$ is a rational point if $l=2$, or at least $4$ if $l\geq 3$. | |
Nov 17, 2018 at 19:59 | history | edited | Vlad Matei | CC BY-SA 4.0 |
Mistake in above
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Nov 17, 2018 at 19:03 | comment | added | jack | Thanks! Do you know if the result is still true for $k=3$? | |
Nov 17, 2018 at 17:44 | comment | added | Vlad Matei | Just found the reference to the problem. It is a problem from AMM, namely Problem E2998 by Clark Kimberling. The problem is stated for the integer version and again four consecutive values suffice. | |
Nov 17, 2018 at 16:24 | history | answered | Vlad Matei | CC BY-SA 4.0 |